3.1591 \(\int \frac{1}{\sqrt{a^2+2 a b x+b^2 x^2}} \, dx\)

Optimal. Leaf size=35 \[ \frac{(a+b x) \log (a+b x)}{b \sqrt{a^2+2 a b x+b^2 x^2}} \]

[Out]

((a + b*x)*Log[a + b*x])/(b*Sqrt[a^2 + 2*a*b*x + b^2*x^2])

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Rubi [A]  time = 0.0081635, antiderivative size = 35, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.1, Rules used = {608, 31} \[ \frac{(a+b x) \log (a+b x)}{b \sqrt{a^2+2 a b x+b^2 x^2}} \]

Antiderivative was successfully verified.

[In]

Int[1/Sqrt[a^2 + 2*a*b*x + b^2*x^2],x]

[Out]

((a + b*x)*Log[a + b*x])/(b*Sqrt[a^2 + 2*a*b*x + b^2*x^2])

Rule 608

Int[1/Sqrt[(a_) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[(b/2 + c*x)/Sqrt[a + b*x + c*x^2], Int[1/(b/2
+ c*x), x], x] /; FreeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rubi steps

\begin{align*} \int \frac{1}{\sqrt{a^2+2 a b x+b^2 x^2}} \, dx &=\frac{\left (a b+b^2 x\right ) \int \frac{1}{a b+b^2 x} \, dx}{\sqrt{a^2+2 a b x+b^2 x^2}}\\ &=\frac{(a+b x) \log (a+b x)}{b \sqrt{a^2+2 a b x+b^2 x^2}}\\ \end{align*}

Mathematica [A]  time = 0.0065614, size = 26, normalized size = 0.74 \[ \frac{(a+b x) \log (a+b x)}{b \sqrt{(a+b x)^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/Sqrt[a^2 + 2*a*b*x + b^2*x^2],x]

[Out]

((a + b*x)*Log[a + b*x])/(b*Sqrt[(a + b*x)^2])

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Maple [A]  time = 0.043, size = 25, normalized size = 0.7 \begin{align*}{\frac{ \left ( bx+a \right ) \ln \left ( bx+a \right ) }{b}{\frac{1}{\sqrt{ \left ( bx+a \right ) ^{2}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((b*x+a)^2)^(1/2),x)

[Out]

(b*x+a)*ln(b*x+a)/b/((b*x+a)^2)^(1/2)

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Maxima [A]  time = 1.0589, size = 19, normalized size = 0.54 \begin{align*} \sqrt{\frac{1}{b^{2}}} \log \left (x + \frac{a}{b}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/((b*x+a)^2)^(1/2),x, algorithm="maxima")

[Out]

sqrt(b^(-2))*log(x + a/b)

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Fricas [A]  time = 1.65035, size = 22, normalized size = 0.63 \begin{align*} \frac{\log \left (b x + a\right )}{b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/((b*x+a)^2)^(1/2),x, algorithm="fricas")

[Out]

log(b*x + a)/b

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Sympy [A]  time = 0.087553, size = 7, normalized size = 0.2 \begin{align*} \frac{\log{\left (a + b x \right )}}{b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/((b*x+a)**2)**(1/2),x)

[Out]

log(a + b*x)/b

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Giac [A]  time = 1.15692, size = 23, normalized size = 0.66 \begin{align*} \frac{\log \left ({\left | b x + a \right |}\right ) \mathrm{sgn}\left (b x + a\right )}{b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/((b*x+a)^2)^(1/2),x, algorithm="giac")

[Out]

log(abs(b*x + a))*sgn(b*x + a)/b